ONE MORE STEP TOWARD BETTER TOMORROW : Complex numbers as Matrices

Tuesday, 26 March 2013

Complex numbers as Matrices

In this section, we use matrices to give a representation of complex numbers. Indeed, consider the set

$\begin{displaymath}{\cal C} = \left\{ \left(\begin{array}{rr} a&b\\ -b&a\\ \en... ...\right); \; \mbox{where $a$ and $b$ are real numbers.}\right\}.\end{displaymath}$

We will write

$\begin{displaymath}M_{a,b} = \left(\begin{array}{rr} a&b\\ -b&a\\ \end{array}\right).\end{displaymath}$

Clearly, the set $\cal C$ is not empty. For example, we have

$\begin{displaymath}M_{0,0} = {\cal O},\;\mbox{and}\; M_{1,0} = I_2.\end{displaymath}$

In particular, we have

$\begin{displaymath}M_{a,b} = M_{c,d}\;\;\mbox{if and only if $a=c$ and $b=d$}\end{displaymath}$

for any real numbers a, b, c, and d.

Algebraic Properties of $\cal C$

1.: Addition: For any real numbers a, b, c, and d, we have

M_a,b + M_c,d = M_a+c,b+d.

In other words, if we add two elements of the set $\cal C$ , we still get a matrix in $\cal C$ . In particular, we have

-M_a,b = M_-a,-b.
2.: Multiplication by a number: We have

$\begin{displaymath}\lambda M_{a,b} = \lambda \left(\begin{array}{rr} a&b\\ -b&a... ...da b&\lambda a\\ \end{array}\right) = M_{\lambda a,\lambda b}.\end{displaymath}$

So a multiplication of an element of $\cal C$ and a number gives a matrix in $\cal C$ .
2.: Multiplication: For any real numbers a, b, c, and d, we have

$\begin{displaymath}M_{a,b} M_{c,d} = \left(\begin{array}{rr} a&b\\ -b&a\\ \end... ...}{rr} ac - bd&ad + bc\\ -bc-ad &ac - bd\\ \end{array}\right).\end{displaymath}$

In other words, we have

M_a,b M_c,d = M_{ac-bd, ad+bc}.

This is an extraordinary formula. It is quite conceivable given the difficult form of the matrix multiplication that, a priori, the product of two elements of $\cal C$ may not be in $\cal C$ again. But, in this case, it turns out to be true.

The above properties infer to $\cal C$ a very nice structure. The next natural question to ask, in this case, is whether a nonzero element of $\cal C$ is invertible. Indeed, for any real numbers a and b, we have

$\begin{displaymath}\left(\begin{array}{rr} a&b\\ -b&a\\ \end{array}\right) \le... ...+ b^2)\left(\begin{array}{rr} 1&0\\ 0&1\\ \end{array}\right).\end{displaymath}$

So, if $a^2 + b^2 \neq 0$ , the matrix M_a,b is invertible and

$\begin{displaymath}M_{a,b}^{-1} = \frac{1}{a^2 + b^2} \left(\begin{array}{rr} a&-b\\ b&a\\ \end{array}\right)= \frac{1}{a^2 + b^2}M_{a,-b} .\end{displaymath}$

In other words, any nonzero element M_a,b of $\cal C$ is invertible and its inverse is still in $\cal C$ since

$\begin{displaymath}M_{a,b}^{-1} = M_{\displaystyle \frac{a}{a^2 + b^2},\frac{b}{a^2 + b^2}}\end{displaymath}$

In order to define the division in $\cal C$ , we will use the inverse. Indeed, recall that

$\begin{displaymath}a \div b = \frac{a}{b} = a \cdot \frac{1}{b} = a \cdot b^{-1}\end{displaymath}$

So for the set $\cal C$ , we have

M_{a, b}÷M_{c, d} = M_{a, b}×M_{c, d}^-1 = M_{a, b}×M_,

which implies

M_{a, b}÷M_{c, d} = M_{, -}

The matrix M_a,-b is called the conjugate of M_a,b. Note that the conjugate of the conjugate of M_a,b is M_a,b itself.

Fundamental Equation. For any M_a,b in $\cal C$ , we have

M_a,b = a M_1,0 + b M_0,1 = a I₂ + b M_0,1.

Note that

M_0,1 M_0,1 = M_-1,0 = - I₂.

Remark. If we introduce an imaginary number i such that i² = -1, then the matrix M_a,b may be rewritten by

a + bi

Tuesday, 26 March 2013

Complex numbers as Matrices

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